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DISCRETE MATHEMATICS

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36. Suppose f (n) is a polynomial such that f(n) is a prime p for some number for all n > 0, Show that f(n) prime p. [Hint: Proof by contradiction.]
36. If the result is false, then f(n) ao + ain + + at nt for some t I. Since f(0) ao = p is prime, f(n) png(n) for g(n) aa2n+.+an Replacing n by pm, we have f(pm) p+ npg(pm). The right hand side is divisible by the prime p, hence f (pm) is divisible by p. But f(pr) is prime, by hypothesis, so f(pn) = p. This means g(pr) = 0, contradicting the fact that a polynomial has only finitely many roots.

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